Ask question

Ask question

All categories

2 years ago

10

A coffee shop uses two different-sized bins to store coffee beans. The volume of the smaller bin is x cubic inches and the volum

e of the larger bin is y cubic inches. The store has S smaller bins and L larger bins. Which expressions represents the total volume of all the bins?

1 answer:

raketka [301]2 years ago

5 0

Answer:

Total volume of all the bins = xS + yL

Step-by-step explanation:

Given: x cubic inches represent the volume of the smaller bin and y cubic inches represents the volume of the larger bin. The store has S smaller bins and L larger bins.

To find: an expression that represents the total volume of all the bins

Solution:

The volume of the smaller bin = x cubic inches

The volume of the larger bin = y cubic inches

Also, the store has S smaller bins and L larger bins

So,

Total volume of all the bins = xS + yL

You might be interested in

Which graph represents the function of f(x) = the quantity of 9 x squared plus 9 x minus 18, all over 3 x plus 6

lora16 [44]

Answer:

The answer is the option C

graph of $3x$ minus $3$ , with discontinuity at negative $2$ , negative $9$

Step-by-step explanation:

we have

$f(x)=\frac{9x^{2}+9x-18}{3x+6}$

Simplify

$f(x)=9\frac{(x^{2}+x-2)}{3(x+2)}$

$f(x)=3\frac{(x^{2}+x-2)}{(x+2)}$

Step 1

Convert to a factored form the numerator

$x^{2}+x-2=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}+x=2$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}+x+0.25=2+0.25$

$x^{2}+x+0.25=2.25$

Rewrite as perfect squares

$(x+0.5)^{2}=2.25$

Square root both sides

$x+0.5=(+/-)1.5$

$x=-0.5(+/-)1.5$

$x=-0.5+1.5=1$

$x=-0.5-1.5=-2$

so

$x^{2}+x-2=(x-1)(x+2)$

Step 2

Simplify the function f(x)

$f(x)=3\frac{(x^{2}+x-2)}{(x+2)}=3\frac{(x-1)(x+2)}{(x+2)}$

The domain of the function f(x) is all real numbers except the number $x=-2$

Because the denominator can not be zero

$f(x)=3\frac{(x-1)(x+2)}{(x+2)}=3(x-1)=3x-3$

$f(x)=3x-3$ ------> with a discontinuity at $x=-2$

$f(-2)=3(-2)-3=-9$

The discontinuity is at point $(-2,-9)$

the answer in the attached figure

8 0

2 years ago

Read 2 more answers

Triangle G E F is shown. Angle G E F is a right angle. The length of hypotenuse F G is 14.5 and the length of F E is 11.9. Angle

jekas [21]

Answer:

B. cos−1(StartFraction 11.9 Over 14.5 EndFraction) = θ

Step-by-step explanation:

From definition:

cos(θ) = adjacent/hypotenuse

The adjacent side respect angle GFE (or θ) is side FE, and side FG is the hypotenuse. Replacing with data and isolating θ:

cos(θ) = 11.9/14.5

θ = cos^-1(11.9/14.5)

3 0

1 year ago

Read 2 more answers

The time needed to travel a certain distance varies inversely with the rate of speed. If it takes 8 hours to travel a certain di

katen-ka-za [31]

I believe it would be 4.8 hours

5 0

1 year ago

Read 2 more answers

A piece of paper is to display ~128~ 128 space, 128, space square inches of text. If there are to be one-inch margins on both si

Grace [21]

Answer:

The dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches

Step-by-step explanation:

We have that:

$Area = 128$

Let the dimension of the paper be x and y;

Such that:

$Length = x$

$Width = y$

So:

$Area = x * y$

Substitute 128 for Area

$128 = x * y$

Make x the subject

$x = \frac{128}{y}$

When 1 inch margin is at top and bottom

The length becomes:

$Length = x + 1 + 1$

$Length = x + 2$

When 2 inch margin is at both sides

The width becomes:

$Width = y + 2 + 2$

$Width = y + 4$

The New Area (A) is then calculated as:

$A = (x + 2) * (y + 4)$

Substitute $\frac{128}{y}$ for x

$A = (\frac{128}{y} + 2) * (y + 4)$

Open Brackets

$A = 128 + \frac{512}{y} + 2y + 8$

Collect Like Terms

$A = \frac{512}{y} + 2y + 8+128$

$A = \frac{512}{y} + 2y + 136$

$A= 512y^{-1} + 2y + 136$

To calculate the smallest possible value of y, we have to apply calculus.

Different A with respect to y

$A' = -512y^{-2} + 2$

Set

$A' = 0$

This gives:

$0 = -512y^{-2} + 2$

Collect Like Terms

$512y^{-2} = 2$

Multiply through by $y^2$

$y^2 * 512y^{-2} = 2 * y^2$

$512 = 2y^2$

Divide through by 2

$256=y^2$

Take square roots of both sides

$\sqrt{256=y^2$

$16=y$

$y = 16$

Recall that:

$x = \frac{128}{y}$

$x = \frac{128}{16}$

$x = 8$

Recall that the new dimensions are:

$Length = x + 2$

$Width = y + 4$

So:

$Length = 8 + 2$

$Length = 10$

$Width = 16 + 4$

$Width = 20$

To double-check;

Differentiate A'

$A' = -512y^{-2} + 2$

$A" = -2 * -512y^{-3}$

$A" = 1024y^{-3}$

$A" = \frac{1024}{y^3}$

The above value is:

$A" = \frac{1024}{y^3} > 0$

This means that the calculated values are at minimum.

<em>Hence, the dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches</em>

3 0

1 year ago

Amy has a box containing 6 white, 4 red, and 8 black marbles. She picks a marble randomly. It is red. The second time, she picks

Alika [10]

8/15 or 11/15 would b the probabiltity of those

6 0

1 year ago